Question

Light of wavelength 5880Å is incident on a thin glass plate (m = 1.5) such that the angle of refraction in the plate is 30⁰. The minimum thickness of the plate, so that it appears dark in the reflected light will be –

• A. 2940 Å
• B. 4074 Å
• C. 2263 Å
• D. 3394 Å

Answer 1

Answer: (C)

2263 Å

Answer 2

For destructive interference

m (2d sec 30⁰) _ 2d tan 30⁰ sin i = nl

n = 1, 2, 3 &&

By snell's law 1 sin i = $\frac{3}{2}$sin30⁰ = $\frac{3}{4}$

\ d = $\frac{2n\lambda}{3\sqrt{3}}$

d_min = $\frac { 2 \lambda } { 3 \sqrt { 3 } }$ ${}_{–}^{\sim}$2263 Å

Altier : 2md cos r = nl for destructive

interference.